Can you give me the comprehensive detailed formula and computation

Certainly! To comprehensively analyze the reliability of the product using the provided data, we can calculate several statistical measures. Here are the steps along with the relevant formulas and computations:

### 1. **Mean (Average)**
The mean is the sum of all the data points divided by the number of data points.

\[
\text{Mean} (\bar{x}) = \frac{\sum_{i=1}^{n} x_i}{n}
\]

Where:
- \( x_i \) are the data points
- \( n \) is the number of data points

**Given Data:**
70.5, 71.2, 67.8, 70.7, 71.1, 69.5, 70.2, 70.6

**Computation:**
\[
\bar{x} = \frac{70.5 + 71.2 + 67.8 + 70.7 + 71.1 + 69.5 + 70.2 + 70.6}{8}
\]
\[
\bar{x} = \frac{561.6}{8} = 70.2
\]

### 2. **Variance (σ²)**
Variance measures how much the data points differ from the mean.

\[
\text{Variance} (\sigma^2) = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}
\]

**Computation:**
1. Calculate the deviations from the mean for each data point:
   - \( 70.5 - 70.2 = 0.3 \)
   - \( 71.2 - 70.2 = 1.0 \)
   - \( 67.8 - 70.2 = -2.4 \)
   - \( 70.7 - 70.2 = 0.5 \)
   - \( 71.1 - 70.2 = 0.9 \)
   - \( 69.5 - 70.2 = -0.7 \)
   - \( 70.2 - 70.2 = 0.0 \)
   - \( 70.6 - 70.2 = 0.4 \)

2. Square each deviation:
   - \( 0.3^2 = 0.09 \)
   - \( 1.0^2 = 1.00 \)
   - \( (-2.4)^2 = 5.76 \)
   - \( 0.5^2 = 0.25 \)
   - \( 0.9^2 = 0.81 \)
   - \( (-0.7)^2 = 0.49 \)
   - \( 0.0^2 = 0.00 \)
   - \( 0.4^2 = 0.16 \)

3. Sum these squared deviations:
   - \( 0.09 + 1.00 + 5.76 + 0.25 + 0.81 + 0.49 + 0.00 + 0.16 = 8.56 \)

4. Divide by the number of data points:
   - \( \sigma^2 = \frac{8.56}{8} = 1.07 \)

### 3. **Standard Deviation (σ)**
The standard deviation is the square root of the variance, showing the average amount of variability in the data.

\[
\text{Standard Deviation} (\sigma) = \sqrt{\sigma^2}
\]

**Computation:**
\[
\sigma = \sqrt{1.07} \approx 1.034
\]

### 4. **Confidence Interval for the Mean**
To estimate the reliability with a certain level of confidence (e.g., 95%), we can construct a confidence interval.

The formula for the confidence interval for the mean is:

\[
\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( t_{\alpha/2} \) is the t-value from the t-distribution table for a 95% confidence level (for \( n-1 = 7 \) degrees of freedom).
- \( \sigma \) is the standard deviation.
- \( n \) is the number of data points.

**Computation:**
1. Look up the t-value for 95% confidence and 7 degrees of freedom, which is approximately 2.3646.
2. Calculate the margin of error:
   \[
   ME = 2.3646 \cdot \frac{1.034}{\sqrt{8}} \approx 0.865
   \]

3. Construct the confidence interval:
   \[
   CI = 70.2 \pm 0.865
   \]
   \[
   CI = (69.335, 71.065)
   \]

### **Summary**
- **Mean:** 70.2
- **Variance:** 1.07
- **Standard Deviation:** 1.034
- **95% Confidence Interval:** (69.335, 71.065)

These computations provide a statistical summary of the sample data, offering insights into the reliability of the insurance product.

Would you like to explore more details or have any specific questions on this?

### Follow-Up Questions:
1. How can we interpret the confidence interval in the context of reliability?
2. What are the implications if the standard deviation is high for reliability?
3. How can we test if the mean is significantly different from a target value?
4. What would be the effect of increasing the sample size on the confidence interval?
5. Can we consider this sample data to be normally distributed based on the information provided?

### Tip:
Understanding how to calculate and interpret the confidence interval is crucial, as it gives a range within which the true mean of the population is expected to lie with a certain level of confidence.

Explore a comprehensive statistical analysis on data reliability, covering mean, variance, standard deviation, and constructing a confidence interval. Detailed computations and interpretations provided with practical examples.

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